1. ## Prove Convergence

1a.) Prove the sequence converges which is defined by x_1 = 3 and x_(n + 1) = 1/(4 - x_n)

b.) Now we know lim x_n exists... explain why lim x_(n + 1) must also exist, and further, explain why it must equal the same value.

c.) Of the recursive equation shown in a.), take the lim of each side of the recursive eq'n. to explicity compute lim x_n

My work:

a.) Now I am not exactly sure how to prove this... I've learned the epsilon method, showing that there exists one greater than 0.. blah blah blah. But I was thinking of maybe trying to use proof by induction to show:

0 <= x_n <= 3 and x_(n) is decreases, which will imply that {x_n} converges (by the monotone convergence thm)...

b.) Not sure..

c.) Maybe something like:

x = 1/(4 - x) => x = 2 - sqrt(3) since 0 <= x <= 3

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Help?

2. Originally Posted by seerTneerGevoLI
1a.) Prove the sequence converges which is defined by x_1 = 3 and x_(n + 1) = 1/(4 - x_n)

b.) Now we know lim x_n exists... explain why lim x_(n + 1) must also exist, and further, explain why it must equal the same value.

c.) Of the recursive equation shown in a.), take the lim of each side of the recursive eq'n. to explicity compute lim x_n

My work:

a.) Now I am not exactly sure how to prove this... I've learned the epsilon method, showing that there exists one greater than 0.. blah blah blah. But I was thinking of maybe trying to use proof by induction to show:

0 <= x_n <= 3 and x_(n) is decreases, which will imply that {x_n} converges (by the monotone convergence thm)...
Yes

b.) Not sure..
Look at the epsilon method and assume that you have the $\displaystyle N_{\varepsilon}$ for the sequence $\displaystyle x_n$, then this will also do for the sequence $\displaystyle x_{n+1}$

c.) Maybe something like:

x = 1/(4 - x) => x = 2 - sqrt(3) since 0 <= x <= 3
Yes

RonL

3. Originally Posted by seerTneerGevoLI
b.) Now we know lim x_n exists... explain why lim x_(n + 1) must also exist, and further, explain why it must equal the same value.
If $\displaystyle (x_n)_{n\in\mathbf{N}}$ is a sequence and $\displaystyle \lim_{n\to\infty}x_n=l$, then any subsequence of the sequence has the same limit.

$\displaystyle (x_{n+1})_{n\in\mathbf{N}}$ is a subsequence of $\displaystyle (x_n)_{n\in\mathbf{N}}$.
$\displaystyle (x_n)_{n\in\mathbf{N}}$ is convergent, so $\displaystyle (x_{n+1})_{n\in\mathbf{N}}$ is also convergent to the same limit.